261

CHAPTER 9

Section 9.1 1.

a. ( ) ( ) ( ) 4.5.41.4 −=−=−=− YEXEYXE , irrespective of sample sizes.

b. ( ) ( ) ( ) ( ) ( )0724.

1000.2

1008.1 222

22

1 =+=+=+=−nm

YVXVYXVσσ

, and the s.d.

of 2691.0724. ==− YX . c. A normal curve with mean and s.d. as given in a and b (because m = n = 100, the CLT

implies that both X and Y have approximately normal distributions, so YX − does also). The shape is not necessarily that of a normal curve when m = n = 10, because the CLT cannot be invoked. So if the two lifetime population distributions are not normal,

the distribution of YX − will typically be quite complicated.

2. The test statistic value is

ns

ms

yxz

22

21 +

−= , and Ho will be rejected if either 96.1≥z or

96.1−≤z . We compute 85.433.433

2100

45

1900

452200

400,40500,4222

==

+

−=z . Since 4.85 >

1.96, reject Ho and conclude that the two brands differ with respect to true average tread lives.

3. The test statistic value is ( )

ns

ms

yxz

22

21

5000

+

−−= , and Ho will be rejected at level .01 if

33.2≥z . We compute ( )

76.193.396

700

45

1500

452200

5000800,36500,4322

==

+

−−=z , which is not

> 2.33, so we don’t reject Ho and conclude that the true average life for radials does not exceed that for economy brand by more than 500.

Chapter 9: Inferences Based on Two Samples

262

4. a. From Exercise 2, the C.I. is

( ) ( ) ( ) 33.849210033.43396.1210096.122

21 ±=±=+±−

ns

ms

yx

( )33.2949,67.1250= . In the context of this problem situation, the interval is moderately wide (a consequence of the standard deviations being large), so the information about 1µ and 2µ is not as precise as might be desirable.

b. From Exercise 3, the upper bound is

( ) 95.635295.652570093.396645.15700 =+=+ . 5.

a. Ha says that the average calorie output for sufferers is more than 1 cal/cm2/min below that

for nonsufferers. ( ) ( )

1414.1016.

1004. 222

221 =+=+

nmσσ

, so

( ) ( )90.2

1414.105.264.

−=−−−

=z . At level .01, Ho is rejected if 33.2−≤z ; since –

2.90 < -2.33, reject Ho.

b. ( ) 0019.90.2 =−Φ=P

c. ( ) 8212.92.11414.

12.133.21 =−Φ−=

+−

−−Φ−=β

d. ( )

( )15.65

2.28.133.22.

2

2

=−

+== nm , so use 66.

Chapter 9: Inferences Based on Two Samples

263

6.

a. Ho should be rejected if 33.2≥z . Since ( )

33.253.3

3296.1

4056.2

87.1612.18≥=

+

−=z , Ho

should be rejected at level .01.

b. ( ) ( ) 3085.50.3539.

0133.21 =−Φ=

−

−Φ=β

c. ( )

06.370529.96.1

1169.28.1645.1

196.14056.2

2=⇒=⇒=

+=+ n

nn, so use

n = 38. d. Since n = 32 is not a large sample, it would no longer be appropriate to use the large

sample test. A small sample t procedure should be used (section 9.2), and the appropriate conclusion would follow.

7.

1 Parameter of interest: =− 21 µµ the true difference of means for males and

females on the Boredom Proneness Rating. Let =1µ men’s average and =2µ women’s average.

2 Ho: 021 =− µµ

3 Ha: 021 >− µµ

4 ( ) ( )

ns

ms

yx

ns

ms

yxz o

22

21

22

21

0

+

−−=

+

∆−−=

5 RR: 645.1≥z

6 ( )

83.1

14868.4

9783.4

26.940.1022

=

+

∆−−= oz

7 Reject Ho. The data indicates the Boredom Proneness Rating is higher for males than for females.

Chapter 9: Inferences Based on Two Samples

264

8. a.

1 Parameter of interest: =− 21 µµ the true difference of mean tensile strength of the

1064 grade and the 1078 grade wire rod. Let =1µ 1064 grade average and =2µ 1078 grade average.

2 Ho: 1021 −=− µµ

3 Ha: 1021 −<− µµ

4 ( ) ( ) ( )

ns

ms

yx

ns

ms

yxz o

22

21

22

21

10

+

−−−=

+

∆−−=

5 RR: α<− valuep

6 ( ) ( )

57.28210.

6

1290.2

1293.1

106.1236.10722

−=−

=

+

−−−=z

7 For a lower-tailed test, the p-value = ( ) 057.28 ≈−Φ , which is less than any α , so reject Ho. There is very compelling evidence that the mean tensile strength of the 1078 grade exceeds that of the 1064 grade by more than 10.

b. The requested information can be provided by a 95% confidence interval for 21 µµ − :

( ) ( ) ( ) ( )588.5,412.6210.96.1696.122

21 −−=±−=+±−

ns

ms

yx .

9.

a. point estimate 2.67.139.19 =−=− yx . It appears that there could be a difference. b.

Ho: 021 =− µµ ,Ha: 021 ≠− µµ , ( )

14.144.52.6

608.15

601.39

7.139.1922

==

+

−=z , and

the p-value = 2[P(z > 1.14)] = 2( .1271) = .2542. The p value is larger than any reasonable α, so we do not reject H0. There is no significant difference.

c. No. With a normal distribution, we would expect most of the data to be within 2 standard deviations of the mean, and the distribution should be symmetric. 2 sd’s above the mean is 98.1, but the distribution stops at zero on the left. The distribution is positively skewed.

d. We will calculate a 95% confidence interval for µ, the true average length of stays for

patients given the treatment. ( )8.21,0.109.99.1960

1.3996.19.19 =±=±

Chapter 9: Inferences Based on Two Samples

265

10. a. The hypotheses are Ho: 521 =− µµ and Ha: 521 >− µµ . At level .001, Ho should

be rejected if 08.3≥z . Since ( )

08.389.22272.

58.596.65<=

−−=z , Ho cannot be

rejected in favor of Ha at this level, so the use of the high purity steel cannot be justified.

b. 121 =∆−− oµµ , so ( ) 2891.53.2272.

108.3 =−Φ=

−Φ=β

11. ( )ns

ms

zYX22

21

2/ +±− α . Standard error = ns

. Substitution yields

( ) ( ) ( )22

212/ SESEzyx +±− α . Using ,05.=α 96.12/ =αz , so

( ) ( ) ( ) ( )41.2,99.02.03.096.18.35.5 22 =+±− . Because we selected ,05.=α we

can state that when using this method with repeated sampling, the interval calculated will bracket the true difference 95% of the time. The interval is fairly narrow, indicating precision of the estimate.

12. The C.I. is ( ) ( ) 46.277.89104.58.277.858.222

21 ±−=±−=+±−

ns

ms

yx

( )31.6,23.11 −−= . With 99% confidence we may say that the true difference between the average 7-day and 28-day strengths is between -11.23 and -6.31 N/mm2.

13. 05.21 == σσ , d = .04, 05.,01. == βα , and the test is one-tailed, so

( )( )38.49

0016.645.133.20025.0025. 2

=++

=n , so use n = 50.

14. The appropriate hypotheses are Ho: 0=θ vs. Ha: 0<θ , where 212 µµθ −= . ( 0<θ is

equivalent to 212 µµ < , so normal is more than twice schizo) The estimator of θ is

YX −= 2θ̂ , with ( ) ( ) ( )nm

YVarXVarVar22

214

4ˆ σσθ +=+= , θσ is the square root

of ( )θ̂Var , and θσ̂ is obtained by replacing each 2iσ with 2

iS . The test statistic is then

θσθ

ˆ

ˆ (since 0=oθ ), and Ho is rejected if .33.2−≤z With ( ) 97.35.669.22ˆ −=−=θ

and ( ) ( )

9236.4503.4

433.24ˆ

22

=+=θσ , 05.19236.

97.−=

−=z ; Because –1.05 > -2.33,

Ho is not rejected.

Chapter 9: Inferences Based on Two Samples

266

15.

a. As either m or n increases, σ decreases, so σµµ o∆−− 21 increases (the numerator is

positive), so

∆−−

−σµµ

αoz 21 decreases, so

∆−−

−Φ=σµµ

β αoz 21

decreases.

b. As β decreases, βz increases, and since βz is the numerator of n , n increases also.

16.

nns

ns

yxz

22.

22

21

=

+

−= . For n = 100, z = 1.41 and p-value = ( )[ ] 1586.41.112 =Φ− .

For n = 400, z = 2.83 and p-value = .0046. From a practical point of view, the closeness of x and y suggests that there is essentially no difference between true average fracture toughness for type I and type I steels. The very small difference in sample averages has been magnified by the large sample sizes – statistical rather than practical significance. The p-value by itself would not have conveyed this message.

Section 9.2 17.

a. ( )

( ) ( )1743.17

44.1694.21.37

99

2

1062

105

2

106

105

22

22

≈=+

=

+

+=ν

b. ( )

( ) ( )217.21

411.694.01.24

149

2

1562

105

2

156

105

22

22

≈=+

=

+

+=ν

c. ( )

( ) ( )1827.18

411.018.84.7

149

2

1562

102

2

156

102

22

22

≈=+

=

+

+=ν

d. ( )

( ) ( )2605.26

098.395.84.12

2311

2

2462

125

2

246

125

22

22

≈=+

=

+

+=ν

Chapter 9: Inferences Based on Two Samples

267

18. With Ho: 021 =− µµ vs. Ha: 021 ≠− µµ , we will reject Ho if α<− valuep .

( )( ) ( )

68.6

45

2

5240.2

6164.

2

5240.

6164.

22

22

≈=

+

+=ν , and the test statistic

17.61265.78.95.2173.22

5240.

6164. 22

==+

−=t leads to a p-value of 2[ P(t > 6.17)] < 2(.0005) =.001,

which is less than most reasonable s'α , so we reject Ho and conclude that there is a difference in the densities of the two brick types.

19. For the given hypotheses, the test statistic 20.1007.3

6.3103.1297.115

638.5

603.5 22

−=−

=+

+−=t , and

the d.f. is ( )

( ) ( )96.9

58241.4

52168.4

8241.42168.422

2

=+

+=ν , so use d.f. = 9. We will reject Ho if

;764.29,01. −=−≤ tt since –1.20 > -2.764, we don’t reject Ho.

20. We want a 95% confidence interval for 21 µµ − . 262.29,025. =t , so the interval is

( ) ( )20.3,40.10007.3262.26.3 −=±− . Because the interval is so wide, it does not appear that precise information is available.

21. Let =1µ the true average gap detection threshold for normal subjects, and =2µ the

corresponding value for CTS subjects. The relevant hypotheses are Ho: 021 =− µµ vs.

Ha: 021 <− µµ , and the test statistic 46.23329.

82.07569.0351125.

53.271.1−=

−=

+−

=t .

Using d.f. ( )

( ) ( )1.15

907569.

70351125.

07569.0351125.22

2

=+

+=ν , or 15, the rejection region is

602.215,01. −=−≤ tt . Since –2.46 is not 602.2−≤ , we fail to reject Ho. We have

insufficient evidence to claim that the true average gap detection threshold for CTS subjects exceeds that for normal subjects.

Chapter 9: Inferences Based on Two Samples

268

22. Let =1µ the true average strength for wire-brushing preparation and let =2µ the average strength for hand-chisel preparation. Since we are concerned about any possible difference between the two means, a two-sided test is appropriate. We test 0: 210 =− µµH vs.

0: 21 ≠− µµaH . We need the degrees of freedom to find the rejection region:

( )( ) ( )

33.141632.0039.

3964.2

1111

2

501.42

1258.1

2

1201.4

1258.1

22

22

=+

=

+

+=ν , which we round down to 14, so we

reject Ho if 145.214,025. =≥ tt . The test statistic is

( ) 159.32442.1

93.313.2320.19

1201.4

1258.1 22

−=−

=+

−=t , which is 145.2−≤ , so we reject Ho and

conclude that there does appear to be a difference between the two population average strengths.

23.

a. Normal plots

Using Minitab to generate normal probability plots, we see that both plots illustrate sufficient linearity. Therefore, it is plausible that both samples have been selected from normal population distributions.

P-Value: 0.344A-Squared: 0.396

Anderson-Darling Normality Tes t

N: 24StDev : 0.444206Average: 1.50833

2.31.81.30.8

.999

.99

.95

.80

.50

.20

.05

.01

.001

Pro

bab

ility

H:

Normal Probability Plot for High Quali ty Fabric

Average: 1. 58750St Dev : 0.530330N: 24

Anderson-Darling Normality Tes tA-Squared: -10.670P-Value: 1. 000

1.0 1.5 2.0 2.5

.001

.01

.05

.20

.50

.80

.95

.99

.999

Pro

babi

lity

P :

Normal Probability Plot for Poor Quality Fabric

Chapter 9: Inferences Based on Two Samples

269

b.

0.5 1.5 2.5

Comparative Box Plot for High Quality and Poor Quality Fabric

QualityPoor

QualityHigh

extensibility (%)

The comparative boxplot does not suggest a difference between average extensibility for the two types of fabrics.

c. We test 0: 210 =− µµH vs. 0: 21 ≠− µµaH . With degrees of freedom

( )5.10

00017906.0433265. 2

==ν , which we round down to 10, and using significance level

.05 (not specified in the problem), we reject Ho if 228.210,025. =≥ tt . The test

statistic is ( )

38.0433265.

08.−=

−=t , which is not 228.2≥ in absolute value, so we

cannot reject Ho. There is insufficient evidence to claim that the true average extensibility differs for the two types of fabrics.

24. A 95% confidence interval for the difference between the true firmness of zero-day apples

and the true firmness of 20-day apples is ( )2039.

2066.

96.474.822

,025. +±− νt . We

calculate the degrees of freedom ( ) ( )

83.30

1919

2039.

2066.

2

2039.2

2066.

222

22=

+

+

=ν , so we use 30 df, and

042.230,025. =t , so the interval is ( ) ( )13.4,43.317142.042.278.3 =± . Thus, with

95% confidence, we can say that the true average firmness for zero-day apples exceeds that of 20-day apples by between 3.43 and 4.13 N.

Chapter 9: Inferences Based on Two Samples

270

25. We calculate the degrees of freedom ( )

( ) ( )95.53

3027

2

318.72

285.5

2

318.7

285.5

22

22

=

+

+=ν , or about 54 (normally

we would round down to 53, but this number is very close to 54 – of course for this large number of df, using either 53 or 54 won’t make much difference in the critical t value) so the

desired confidence interval is ( ) 318.7

285.5 22

68.13.885.91 +±−

( )131.6,269.931.22.3 =±= . Because 0 does not lie inside this interval, we can be

reasonably certain that the true difference 21 µµ − is not 0 and, therefore, that the two population means are not equal. For a 95% interval, the t value increases to about 2.01 or so, which results in the interval 506.32.3 ± . Since this interval does contain 0, we can no longer conclude that the means are different if we use a 95% confidence interval.

26. Let =1µ the true average potential drop for alloy connections and let =2µ the true average potential drop for EC connections. Since we are interested in whether the potential drop is higher for alloy connections, an upper tailed test is appropriate. We test 0: 210 =− µµH

vs. 0: 21 >− µµaH . Using the SAS output provided, the test statistic, when assuming

unequal variances, is t = 3.6362, the corresponding df is 37.5, and the p-value for our upper

tailed test would be ½ (two-tailed p-value) = ( ) 0004.0008.21 = . Our p-value of .0004 is

less than the significance level of .01, so we reject Ho. We have sufficient evidence to claim that the true average potential drop for alloy connections is higher than that for EC connections.

27. The approximate degrees of freedom for this estimate are

( )( ) ( )

83.8175.10159.893

75

2

83.82

63.11

2

83.8

63.11

22

22

==

+

+=ν , which we round down to 8, so 306.28,025. =t

and the desired interval is ( ) ( )4674.5306.29.18306.24.213.40 83.8

63.11 22

±=+±−

( )5.31,3.6607.129.18 =±= . Because 0 is not contained in this interval, there is strong

evidence that 21 µµ − is not 0; i.e., we can conclude that the population means are not equal.

Calculating a confidence interval for 12 µµ − would change only the order of subtraction of the sample means, but the standard error calculation would give the same result as before. Therefore, the 95% interval estimate of 12 µµ − would be ( -31.5, -6.3), just the negatives of the endpoints of the original interval. Since 0 is not in this interval, we reach exactly the same conclusion as before; the population means are not equal.

Chapter 9: Inferences Based on Two Samples

271

28. We will test the hypotheses: 10: 210 =− µµH vs. 10: 21 >− µµaH . The test

statistic is ( )( )

08.217.25.410

544.4

1075.2 22

==+

−−=

yxt The degrees of freedom

( )( ) ( )

659.595.308.22

49

2

544.42

1075.2

2

544.4

1075.2

22

22

≈==

+

+=ν and the p-value from table A.8 is approx .04,

which is < .10 so we reject H0 and conclude that the true average lean angle for older females is more than 10 degrees smaller than that of younger females.

29. Let =1µ the true average compression strength for strawberry drink and let =2µ the true average compression strength for cola. A lower tailed test is appropriate. We test

0: 210 =− µµH vs. 0: 21 <− µµaH . The test statistic is

10.2154.29

14−=

+−

=t . ( )

( ) ( )3.25

8114.7736.1971

1415

144.29

4.4422

2

==+

=ν , so use df=25.

The p-value 023.)10.2( =−<≈ tP . This p-value indicates strong support for the

alternative hypothesis. The data does suggest that the extra carbonation of cola results in a higher average compression strength.

30.

a. We desire a 99% confidence interval. First we calculate the degrees of freedom:

( )( ) ( )

24.37

2626

2

263.42

262.2

2

263.4

262.2

22

22

=

+

+=ν , which we would round down to 37, except that there is

no df = 37 row in Table A.5. Using 36 degrees of freedom (a more conservative choice),

719.236,005. =t , and the 99% C.I. is

( ) ( )83.6,98.11576.24.9719.28.424.33 263.4

262.2 22

−−=±−=+±− . We are

very confident that the true average load for carbon beams exceeds that for fiberglass beams by between 6.83 and 11.98 kN.

b. The upper limit of the interval in part a does not give a 99% upper confidence bound. The 99% upper bound would be ( ) 09.79473.434.24.9 −=+− , meaning that the true average load for carbon beams exceeds that for fiberglass beams by at least 7.09 kN.

Chapter 9: Inferences Based on Two Samples

272

31. a.

The mo st notable feature of these boxplots is the larger amount of variation present in the mid-range data compared to the high-range data. Otherwise, both look reasonably symmetric with no outliers present.

b. Using df = 23, a 95% confidence interval for rangehighrangemid −− − µµ is

( ) ( )54.9,84.769.885.069.245.4373.438 1183.6

171.15 22

−=±=+±− . Since

plausible values for rangehighrangemid −− − µµ are both positive and negative (i.e., the

interval spans zero) we would conclude that there is not sufficient evidence to suggest that the average value for mid-range and the average value for high-range differ.

32. Let =1µ the true average proportional stress limit for red oak and let =2µ the true average

proportional stress limit for Douglas fir. We test 1: 210 =− µµH vs. 1: 21 >− µµaH .

The test statistic is ( )

818.12084.83.1165.648.8

1028.1

1479. 22

=+

−−=t . With degrees of freedom

( )( ) ( )

1485.13

913

2084.2

1028.12

1479.

2

22≈=

+

=ν , the p-value 046.)8.1( =>≈ tP . This p-value

indicates strong support for the alternative hypothesis since we would reject Ho at significance levels greater than .046. There is sufficient evidence to claim that true average proportional stress limit for red oak exceeds that of Douglas fir by more than 1 MPa.

high rangem id range

470

460

450

440

430

420m

id r

ange

Comparative Box Plot for High Range and Mid Range

Chapter 9: Inferences Based on Two Samples

273

33. Let =1µ the true average weight gain for steroid treatment and let =2µ the true average weight gain for the population not treated with steroids. The exercise asks if we can conclude that 2µ exceeds 1µ by more than 5 g., which we can restate in the equivalent form:

521 −<− µµ . Therefore, we conduct a lower-tailed test of 5: 210 −=− µµH vs.

5: 21 −<− µµaH . The test statistic is

( ) ( ) ( )2.223.2

2124.17.2

105.2

86.2

55.408.32222

221

≈−=−

=

+

−−−=

+

∆−−=

ns

ms

yxt . The approximate d.f. is

( )( ) ( )

876.141454.1609.2

97

2

105.22

86.2

2

105.2

86.2

22

22

==

+

+=ν , which we round down to 14. The p-value for a

lower tailed test is P( t < -2.2 ) = P( t > 2.2 ) = .022. Since this p-value is larger than the specified significance level .01, we cannot reject Ho. Therefore, this data does not support the belief that average weight gain for the control group exceeds that of the steroid group by more than 5 g.

34.

a. Following the usual format for most confidence intervals: statistic ± (critical value)(standard error), a pooled variance confidence interval for the difference between

two means is ( ) nmpnm styx 112,2/ +⋅±− −+α .

b. The sample means and standard deviations of the two samples are 90.13=x ,

225.11 =s , 20.12=y , 010.12 =s . The pooled variance estimate is =2ps

( ) ( )2222

21 010.1

24414

225.1244

142

12

1

−+−

+

−+−

=

−+−

+

−+−

snm

ns

nmm

260.1= , so 1227.1=ps . With df = m+n-1 = 6 for this interval, 447.26,025. =t and

the desired interval is ( ) ( )( ) 41

411227.1447.220.1290.13 +±−

( )64.3,24.943.17.1 −=±= . This interval contains 0, so it does not support the conclusion that the two population means are different.

c. Using the two-sample t interval discussed earlier, we use the CI as follows: First, we need

to calculate the degrees of freedom. ( )

( ) ( )919.9

0686.6302.

33

2

401.12

4225.1

2

401.1

4225.1

22

22

≈==

+

+=ν so

262.29,025. =t . Then the interval is

( ) ( ) ( )50.3,10.7938.262.270.1262.22.129.13 401.1

4225.1 22

−=±=+±− . This

interval is slightly smaller, but it still supports the same conclusion.

Chapter 9: Inferences Based on Two Samples

274

35. There are two changes that must be made to the procedure we currently use. First, the

equation used to compute the value of the t test statistic is: ( ) ( )

nms

yxt

p11

+

∆−−= where sp is

defined as in Exercise 34 above. Second, the degrees of freedom = m + n – 2. Assuming equal variances in the situation from Exercise 33, we calculate sp as follows:

( ) ( ) 544.25.2169

6.2167 22 =

+

=ps . The value of the test statistic is, then,

( ) ( )2.224.2

101

81

544.2

55.408.32−≈−=

+

−−−=t . The degrees of freedom = 16, and the p-

value is P ( t < -2.2) = .021. Since .021 > .01, we fail to reject Ho. This is the same conclusion reached in Exercise 33.

Section 9.3 36. 25.7=d , 8628.11=Ds

1 Parameter of Interest: =Dµ true average difference of breaking load for fabric in unabraded or abraded condition.

2 0:0 =DH µ

3 0: >DaH µ

4 ns

dns

dt

DD

D

/0

/−

=−

=µ

5 RR: 998.27,01. =≥ tt

6 73.18/8628.11

025.7=

−=t

7 Fail to reject Ho. The data does not indicate a difference in breaking load for the two fabric load conditions.

Chapter 9: Inferences Based on Two Samples

275

37. a. This exercise calls for paired analysis. First, compute the difference between indoor and

outdoor concentrations of hexavalent chromium for each of the 33 houses. These 33

differences are summarized as follows: n = 33, 4239.−=d , 3868.=ds , where d =

(indoor value – outdoor value). Then 037.232,025. =t , and a 95% confidence interval

for the population mean difference between indoor and outdoor concentration is

( ) ( )2868.,5611.13715.4239.33

3868.037.24239. −−=±−=

±− . We can be

highly confident, at the 95% confidence level, that the true average concentration of hexavalent chromium outdoors exceeds the true average concentration indoors by between .2868 and .5611 nanograms/m3.

b. A 95% prediction interval for the difference in concentration for the 34th house is

( ) ( )( ) ( )3758,.224.113868.037.24239.1 3311

32,025. −=+±−=+± ndstd .

This prediction interval means that the indoor concentration may exceed the outdoor concentration by as much as .3758 nanograms/m3 and that the outdoor concentration may exceed the indoor concentration by a much as 1.224 nanograms/m3, for the 34th house. Clearly, this is a wide prediction interval, largely because of the amount of variation in the differences.

38.

a. The median of the “Normal” data is 46.80 and the upper and lower quartiles are 45.55 and 49.55, which yields an IQR of 49.55 – 45.55 = 4.00. The median of the “High” data is 90.1 and the upper and lower quartiles are 88.55 and 90.95, which yields an IQR of 90.95 – 88.55 = 2.40. The most significant feature of these boxplots is the fact that their locations (medians) are far apart.

Normal :High:

90

80

70

60

50

40

Comparative Boxplots

for Normal and High Strength Concrete Mix

Chapter 9: Inferences Based on Two Samples

276

b. This data is paired because the two measurements are taken for each of 15 test conditions. Therefore, we have to work with the differences of the two samples. A quantile of the 15 differences shows that the data follows (approximately) a straight line, indicating that it is reasonable to assume that the differences follow a normal distribution. Taking

differences in the order “Normal” – “High” , we find 23.42−=d , and 34.4=ds .

With 145.214,025. =t , a 95% confidence interval for the difference between the

population means is

( ) ( )83.39,63.44404.223.421534.4

145.223.42 −−=±−=

±− . Because 0 is

not contained in this interval, we can conclude that the difference between the population means is not 0; i.e., we conclude that the two population means are not equal.

39.

a. A normal probability plot shows that the data could easily follow a normal distribution.

b. We test 0:0 =dH µ vs. 0: ≠daH µ , with test statistic

7.274.214/228

02.167/

0≈=

−=

−=

nsd

tD

. The two-tailed p-value is 2[ P( t > 2.7)] =

2[.009] = .018. Since .018 < .05, we can reject Ho . There is strong evidence to support the claim that the true average difference between intake values measured by the two methods is not 0. There is a difference between them.

40.

a. Ho will be rejected in favor of Ha if either 947.215,005. =≥ tt or 947.2−≤t . The

summary quantities are 544.−=d , and 714.=ds , so 05.31785.

544.−=

−=t .

Because 947.205.3 −≤− , Ho is rejected in favor of Ha.

b. 31.72 =ps , 70.2=ps , and 57.96.544.

−=−

=t , which is clearly insignificant; the

incorrect analysis yields an inappropriate conclusion.

41. We test 0:0 =dH µ vs. 0: >daH µ . With 600.7=d , and 178.4=ds ,

9.187.139.16.2

9/178.45600.7

≈==−

=t . With degrees of freedom n – 1 = 8, the

corresponding p-value is P( t > 1.9 ) = .047. We would reject Ho at any alpha level greater than .047. So, at the typical significance level of .05, we would (barely) reject Ho, and conclude that the data indicates that the higher level of illumination yields a decrease of more than 5 seconds in true average task completion time.

Chapter 9: Inferences Based on Two Samples

277

42. 1 Parameter of interest: dµ denotes the true average difference of spatial ability in

brothers exposed to DES and brothers not exposed to DES. Let

.expexp osedunosedd µµµ −=

2 0:0 =DH µ

3 0: <DaH µ

4 ns

dns

dt

DD

D

/0

/−

=−

=µ

5 RR: P-value < .05, df = 8

6 ( )

2.25.0

07.136.12−=

−−=t , with corresponding p-value .029 (from Table A.8)

7 Reject Ho. The data supports the idea that exposure to DES reduces spatial ability. 43.

a. Although there is a “jump” in the middle of the Normal Probability plot, the data follow a reasonably straight path, so there is no strong reason for doubting the normality of the population of differences.

b. A 95% lower confidence bound for the population mean difference is:

( ) 14.4954.1060.381518.23

761.160.3814,05. −=−−=

−−=

−

n

std d .

Therefore, with a confidence level of 95%, the population mean difference is above (–49.14).

c. A 95% upper confidence bound for the corresponding population mean difference is

14.4954.1060.38 =+ 44. We need to check the differences to see if the assumption of normality is plausible. A

probability chart will validate our use of the t distribution. A 95% confidence interval:

( ) 91.22263.263516645.508

753.163.263515,05. +=

+=

+

n

std d

( )54.2858,∞⇒ 45. The differences (white – black) are –7.62, -8.00, -9.09, -6.06, -1.39, -16.07, -8.40, -8.89, and

–2.88, from which 600.7−=d , and 178.4=ds . The confidence level is not specified in

the problem description; for 95% confidence, 306.28,025. =t , and the C.I. is

( ) ( )389.4,811.10211.3600.79

178.4306.2600.7 −−=±−=

±− .

46. With ( ) ( )5,6, 11 =yx , ( ) ( )14,15, 22 =yx , ( ) ( )0,1, 33 =yx , and ( ) ( )20,21, 44 =yx ,

1=d and 0=ds (the d I’s are 1, 1, 1, and 1), while s1 = s2 = 8.96, so sp = 8.96 and t = .16.

Chapter 9: Inferences Based on Two Samples

278

Section 9.4 47. Ho will be rejected if 33.201. −=−≤ zz . With 150.ˆ1 =p , and 300.ˆ 2 =p ,

263.800210

6002008030ˆ ==

++

=p , and 737.ˆ =q . The calculated test statistic is

( )( )( )18.4

0359.150.

737.263.

300.150.

6001

2001

−=−

=+

−=z . Because 33.218.4 −≤− , Ho is

rejected; the proportion of those who repeat after inducement appears lower than those who repeat after no inducement.

48.

a. Ho will be rejected if 96.1≥z . With 2100.30063ˆ1 ==p , and 4167.

18075ˆ 2 ==p ,

2875.1803007563ˆ =

++

=p , ( )( )( )

84.40427.2067.

7125.2875.

4167.2100.

1801

3001

−=−

=+

−=z .

Since 96.184.4 −≤− , Ho is rejected. b. 275.=p and 0432.=σ , so power =

( )( )[ ] ( )( )[ ]=

+−

Φ−

+

Φ−0432.

2.0421.96.10432.

2.0421.96.11

( ) ( )[ ] 9967.72.254.61 =Φ−Φ− .

49. 1 Parameter of interest: p1 – p2 = true difference in proportions of those responding to

two different survey covers. Let p1 = Plain, p2 = Picture. 2 0: 210 =− ppH

3 0: 21 <− ppH a

4 ( )nmqp

ppz

11

21

ˆˆ

ˆˆ

+

−=

5 Reject Ho if p-value < .10

6 ( )( )( )

1910.213

12071

420207

420213

213109

207104

−=+

−=z ; p-value = .4247

7 Fail to Reject Ho. The data does not indicate that plain cover surveys have a lower response rate.

Chapter 9: Inferences Based on Two Samples

279

50. Let 05.=α . A 95% confidence interval is ( ) ( )nqp

mqpzpp 2211 ˆˆˆˆ

2/21 ˆˆ +±− α

( ) ( )( ) ( )( ) ( )1708,.0160.0774.0934.266395

96.1 266140

266126

395171

395224

266126

395224 =±=

+±−= .

51.

a. 210 : ppH = will be rejected in favor of 21: ppH a ≠ if either 645.1≥z or

645.1−≤z . With 193.ˆ1 =p , and 182.ˆ 2 =p , 188.ˆ =p , 48.100742.

011.==z .

Since 1.48 is not 645.1≥ , Ho is not rejected and we conclude that no difference exists. b. Using formula (9.7) with p1 = .2, p2 = .18, 1.=α , 1.=β , and 645.12/ =αz ,

( )( )( )6582

0004.1476.16.28.162.138.5.645.1

2

=++

=n

52. Let p1 = true proportion of irradiated bulbs that are marketable; p2 = true proportion of

untreated bulbs that are marketable; The hypotheses are 0: 210 =− ppH vs.

0: 210 >− ppH . The test statistic is ( )nmqp

ppz

11

21

ˆˆ

ˆˆ

+

−= . With 850.

180153ˆ1 ==p , and

661.180119ˆ 2 ==p , 756.

360272ˆ ==p ,

( )( )( )2.4

045.189.

244.756.

661.850.

1801

1801

==+

−=z .

The p-value = ( ) 02.41 ≈Φ− , so reject Ho at any reasonable level. Radiation appears to be beneficial.

53.

a. A 95% large sample confidence interval formula for ( )θln is

( )ny

ynmx

xmz

−+

−± 2/

ˆln αθ . Taking the antilogs of the upper and lower bounds

gives the confidence interval for θ itself.

b. 818.1ˆ037,11

104

034,11189

==θ , ( ) 598.ˆln =θ , and the standard deviation is

( )( ) ( )( ) 1213.104037,11

933,10189034,11

845,10=+ , so the CI for ( )θln is

( ) ( )836,.360.1213.96.1598. =± . Then taking the antilogs of the two bounds gives

the CI for θ to be ( )31.2,43.1 .

Chapter 9: Inferences Based on Two Samples

280

54. a. The “after” success probability is p1 + p3 while the “before” probability is p1 + p2 , so p1 +

p3 > p1 + p2 becomes p3 > p2; thus we wish to test 230 : ppH = versus

23: ppH a > .

b. The estimator of (p1 + p3) – (p1 + p2) is ( ) ( )

nXX

nXXXX 232131 −

=+−+

.

c. When Ho is true, p2 = p3, so n

ppn

XXVar 3223 +

=

−

, which is estimated by

npp 32 ˆˆ +

. The Z statistic is then 32

23

32

23

ˆˆ XXXX

npp

nXX

+−

=+

−

.

d. The computed value of Z is 68.2150200

150200=

+−

, so ( ) 0037.68.21 =Φ−=P . At

level .01, Ho can be rejected but at level .001 Ho would not be rejected.

55. 550.40

715ˆ1 =+

=p , 690.4229ˆ 2 ==p , and the 95% C.I. is

( ) ( ) ( )07,.35.21.14.106.96.1690.550. −=±−=±− .

56. Using p1 = q1 = p2 = q2 = .5, ( )nnn

L 7719.225.25.96.12 =

+= , so L=.1 requires n=769.

Section 9.5 57.

a. From Table A.9, column 5, row 8, 69.38,5,01. =F .

b. From column 8, row 5, 82.45,8,01. =F .

c. 207.1

5,8,05.8,5,95. ==

FF .

Chapter 9: Inferences Based on Two Samples

281

d. 271.1

8,5,05.5,8,95. ==

FF

e. 30.412,10,01. =F

f. 212.71.411

10,12,01.12,10,99. ===

FF .

g. 16.64,6,05. =F , so ( ) 95.16.6 =≤FP .

h. Since 177.64.51

5,10,99. ==F ,

( ) ( ) ( )177.74.474.4177. ≤−≤=≤≤ FPFPFP 94.01.95. =−= . 58.

a. Since the given f value of 4.75 falls between 33.310,5,05. =F and 64.510,5,01. =F , we

can say that the upper-tailed p-value is between .01 and .05. b. Since the given f of 2.00 is less than 52.210,5,10. =F , the p-value > .10.

c. The two tailed p-value = ( ) 02.)01(.264.52 ==≥FP . d. For a lower tailed test, we must first use formula 9.9 to find the critical values:

3030.1

5,10,10.10,5,90. ==

FF , 2110.

1

5,10,05.10,5,95. ==

FF ,

0995.1

5,10,01.10,5,99. ==

FF . Since .0995 < f = .200 < .2110, .01 < p-value < .05 (but

obviously closer to .05). e. There is no column for numerator d.f. of 35 in Table A.9, however looking at both df =

30 and df = 40 columns, we see that for denominator df = 20, our f value is between F.01 and F.001. So we can say .001< p-value < .01.

Chapter 9: Inferences Based on Two Samples

282

59. We test 22

0 21: σσ =H vs.

2221

: σσ ≠aH . The calculated test statistic is

( )( )

384.44.475.2

2

2

==f . With numerator d.f. = m – 1 = 10 – 1 = 9, and denominator d.f. = n –

1 = 5 – 1 = 4, we reject H0 if 00.64,9,05. =≥ Ff or

275.63.311

9,4,05.4,9,95. ===≤ FFf . Since .384 is in neither rejection region, we do

not reject H0 and conclude that there is no significant difference between the two standard deviations.

60. With =1σ true standard deviation for not-fused specimens and =2σ true standard

deviation for fused specimens, we test 210 : σσ =H vs. 21: σσ >aH . The calculated

test statistic is ( )( )

814.19.2053.277

2

2

==f . With numerator d.f. = m – 1 = 10 – 1 = 9, and

denominator d.f. = n – 1 = 8 – 1 = 7, 7,9,10.72.2814.1 Ff =<= . We can say that the p-

value > .10, which is obviously > .01, so we cannot reject Ho. There is not sufficient evidence that the standard deviation of the strength distribution for fused specimens is smaller than that of not-fused specimens.

61. Let =21σ variance in weight gain for low-dose treatment, and =2

2σ variance in weight

gain for control condition. We wish to test 22

210 : σσ =H vs. 2

221: σσ >aH . The test

statistic is 22

21

ss

f = , and we reject Ho at level .05 if 08.222,19,05. ≈> Ff .

( )( )

8.2085.23254

2

2

≥==f , so reject Ho at level .05. The data does suggest that there is

more variability in the low-dose weight gains.

62. 210 : σσ =H will be rejected in favor of 21: σσ ≠aH if either 56.44,47,975. ≈≤ Ff

or if 8.144,47,025. ≈≥ Ff . Because 22.1=f , Ho is not rejected. The data does not

suggest a difference in the two variances.

Chapter 9: Inferences Based on Two Samples

283

63. ασσ

αα −=

≤≤ −−−−− 1

//

1,1,2/22

22

21

21

1,1,2/1 nmnm FSS

FP . The set of inequalities inside the

parentheses is clearly equivalent to 2

1

1,1,2/22

21

22

21

1,1,2/122

S

FS

S

FS nmnm −−−−− ≤≤ αα

σσ

. Substituting

the sample values 21s and 2

2s yields the confidence interval for 21

22

σσ

, and taking the square

root of each endpoint yields the confidence interval for 1

2

σσ

. m = n = 4, so we need

28.93,3,05. =F and 108.28.91

3,3,95. ==F . Then with s1 = .160 and s2 = .074, the C. I.

for 21

22

σσ

is (.023, 1.99), and for 1

2

σσ

is (.15, 1.41).

64. A 95% upper bound for 1

2

σσ

is ( ) ( )

( )10.8

79.

18.359.32

2

21

9,9,05.22 ==

s

Fs. We are

confident that the ratio of the standard deviation of triacetate porosity distribution to that of the cotton porosity distribution is at most 8.10.

Supplementary Exercises 65. We test 0: 210 =− µµH vs. 0: 21 ≠− µµaH . The test statistic is

( ) ( )22.3

524.1550

24150

1041

1027

757807222

221

===

+

−=

+

∆−−=

ns

ms

yxt . The approximate d.f. is

( )( ) ( )

6.15

91.168

99.72

24122

2

=+

=ν , which we round down to 15. The p-value for a two-

tailed test is approximately 2P( t > 3.22) = 2( .003) = .006. This small of a p-value gives strong support for the alternative hypothesis. The data indicates a significant difference.

Chapter 9: Inferences Based on Two Samples

284

66. a.

Although the median of the fertilizer plot is higher than that of the control plots, the fertilizer plot data appears negatively skewed, while the opposite is true for the control plot data.

b. A test of 0: 210 =− µµH vs. 0: 21 ≠− µµaH yields a t value of -.20, and a two-

tailed p-value of .85. (d.f. = 13). We would fail to reject Ho; the data does not indicate a significant difference in the means.

c. With 95% confidence we can say that the true average difference between the tree density

of the fertilizer plots and that of the control plots is somewhere between –144 and 120. Since this interval contains 0, 0 is a plausible value for the difference, which further supports the conclusion based on the p-value.

67. Let p1 = true proportion of returned questionnaires that included no incentive; p2 = true

proportion of returned questionnaires that included an incentive. The hypotheses are

0: 210 =− ppH vs. 0: 210 <− ppH . The test statistic is ( )nmqp

ppz

11

21

ˆˆ

ˆˆ

+

−= .

682.11075ˆ1 ==p , and 673.

9866ˆ 2 ==p . At this point we notice that since 21 ˆˆ pp > , the

numerator of the z statistic will be > 0, and since we have a lower tailed test, the p-value will be > .5. We fail to reject Ho. This data does not suggest that including an incentive increases the likelihood of a response.

Fe rtiliz Co ntrol

1000

1100

1200

1300

1400

Fer

tiliz

Comparative Boxplot of Tree Density BetweenFertilizer Plots and Control Plots

Chapter 9: Inferences Based on Two Samples

285

68. Summary quantities are m = 24, 66.103=x , s1 = 3.74, n = 11, 11.101=y , s2 = 3.60. We

use the pooled t interval based on 24 + 11 – 2 = 33 d.f.; 95% confidence requires

03.233,025. =t . With 68.132 =ps and 70.3=ps , the confidence interval is

( )( ) ( )28.5,18.73.255.270.303.255.2 111

241 −=±=+± . We are confident that the

difference between true average dry densities for the two sampling methods is between -.18 and 5.28. Because the interval contains 0, we cannot say that there is a significant difference between them.

69. The center of any confidence interval for 21 µµ − is always 21 xx − , so

3.6092

9.16913.47321 =

+−=− xx . Furthermore, half of the width of this interval is

( )6.1082

23.4739.1691

=−−

. Equating this value to the expression on the right of the

95% confidence interval formula, ( )2

22

1

2196.16.1082

ns

ns

+= , we find

35.55296.1

6.1082

2

22

1

21 ==+

ns

ns

. For a 90% interval, the associated z value is 1.645, so

the 90% confidence interval is then ( )( ) 6.9083.60935.552645.13.609 ±=±

( )9.1517,3.299−= . 70.

a. A 95% lower confidence bound for the true average strength of joints with a side coating

is ( ) 78.5945.323.631096.5

833.123.639,025. =−=

−=

−

ns

tx . That is,

with a confidence level of 95%, the average strength of joints with a side coating is at least 59.78 (Note: this bound is valid only if the distribution of joint strength is normal.)

b. A 95% lower prediction bound for the strength of a single joint with a side coating is

( ) ( )( )1011

9,025. 196.5833.123.631 +−=+− nstx 77.5146.1123.63 =−= .

That is, with a confidence level of 95%, the strength of a single joint with a side coating would be at least 51.77.

c. For a confidence level of 95%, a two-sided tolerance interval for capturing at least 95%

of the strength values of joints with side coating is ±x (tolerance critical value)s. The tolerance critical value is obtained from Table A.6 with 95% confidence, k = 95%, and n = 10. Thus, the interval is

( )( ) ( )37.83,09.4314.2023.6396.5379.323.63 =±=± . That is, we can be highly confident that at least 95% of all joints with side coatings have strength values between 43.09 and 83.37.

Chapter 9: Inferences Based on Two Samples

286

d. A 95% confidence interval for the difference between the true average strengths for the

two types of joints is ( ) ( ) ( )1096.5

1059.9

23.6395.8022

,025. +±− νt . The

approximate degrees of freedom is ( )

( ) ( )05.15

99

2105216.352

109681.91

2105216.35

109681.91

=

+

+=ν , so we use 15

d.f., and 131.215,025. =t . The interval is , then,

( )( ) ( )33.25,11.1061.772.1757.3131.272.17 =±=± . With 95% confidence, we can say that the true average strength for joints without side coating exceeds that of joints with side coating by between 10.11 and 25.33 lb-in./in.

71. m = n = 40, 0.3975=x , s1 = 245.1, 0.2795=y , s2 = 293.7. The large sample 99%

confidence interval for 21 µµ − is ( )40

7.29340

1.24558.20.27950.3975

22

+±−

( ) ( )1336,10245.15600.1180 ≈± . The value 0 is not contained in this interval so we can

state that, with very high confidence, the value of 21 µµ − is not 0, which is equivalent to

concluding that the population means are not equal. 72. This exercise calls for a paired analysis. First compute the difference between the amount of

cone penetration for commutator and pinion bearings for each of the 17 motors. These 17

differences are summarized as follows: n = 17, 18.4−=d , 85.35=ds , where d =

(commutator value – pinion value). Then 120.216,025. =t , and the 95% confidence interval

for the population mean difference between penetration for the commutator armature bearing and penetration for the pinion bearing is:

( ) ( )25.14,61.2243.1818.41785.35

120.218.4 −=±−=

±− . We would have to say

that the population mean difference has not been precisely estimated. The bound on the error of estimation is quite large. In addition, the confidence interval spans zero. Because of this, we have insufficient evidence to claim that the population mean penetration differs for the two types of bearings.

Chapter 9: Inferences Based on Two Samples

287

73. Since we can assume that the distributions from which the samples were taken are normal, we use the two-sample t test. Let 1µ denote the true mean headability rating for aluminum killed

steel specimens and 2µ denote the true mean headability rating for silicon killed steel. Then

the hypotheses are 0: 210 =− µµH vs. 0: 21 ≠− µµaH . The test statistic is

25.2086083.

66.047203.03888.

66.−=

−=

+−

=t . The approximate degrees of freedom

( )( ) ( )

5.57

29047203.

2903888.

086083.22

2

=+

=ν , so we use 57. The two-tailed p-value

( ) 028.014.2 =≈ , which is less than the specified significance level, so we would reject Ho. The data supports the article’s authors’ claim.

74. Let 1µ denote the true average tear length for Brand A and let 2µ denote the true average

tear length for Brand B. The relevant hypotheses are 0: 210 =− µµH vs.

0: 21 >− µµaH . Assuming both populations have normal distributions, the two-sample t

test is appropriate. m = 16, 0.74=x , s1 = 14.8, n = 14, 0.61=y , s2 = 12.5, so the

approximate d.f. is ( )

( ) ( )97.27

1315

2

145.122

168.14

2

145.12

168.14

22

22

=

+

+=ν , which we round down to 27. The test

statistic is 6.20.610.74

145.12

168.14 22

≈+

−=t . From Table A.7, the p-value = P( t > 2.6) = .007. At a

significance level of .05, Ho is rejected and we conclude that the average tear length for Brand A is larger than that of Brand B.

75.

a. The relevant hypotheses are 0: 210 =− µµH vs. 0: 21 ≠− µµaH . Assuming

both populations have normal distributions, the two-sample t test is appropriate. m = 11, 1.98=x , s1 = 14.2, n = 15, 2.129=y , s2 = 39.1. The test statistic is

84.2252.1201.31

9207.1013309.181.31

−=−

=+

−=t . The approximate degrees of

freedom ( )

( ) ( )64.18

149207.101

103309.18

252.12022

2

=+

=ν , so we use 18. From Table A.7,

the two-tailed p-value ( ) 012.006.2 =≈ . No, obviously, the results are different.

Chapter 9: Inferences Based on Two Samples

288

b. For the hypotheses 25: 210 −=− µµH vs. 25: 21 −<− µµaH , the test statistic

changes to ( )

556.252.120

251.31−=

−−−=t . With degrees of freedom 18, the p-value

( ) 278.6. =−<≈ tP . Since the p-value is greater than any sensible choice of α , we

fail to reject Ho. There is insufficient evidence that the true average strength for males exceeds that for females by more than 25N.

76.

a. The relevant hypotheses are 0: 210 =− ∗∗ µµH (which is equivalent to saying

021 =− µµ ) versus 0: 21 ≠− ∗∗ µµaH (which is the same as saying

021 ≠− µµ ). The pooled t test is based on d.f. = m + n – 2 = 8 + 9 – 2 = 15. The

pooled variance is =2ps 2

221 2

12

1s

nmn

snm

m

−+−

+

−+−

( ) ( )22 6.4298

199.4

29818

−+−

+

−+−

49.22= , so 742.4=ps . The test statistic

is 0.304.3742.4

0.110.18**

91

8111

≈=+

−=

+

−=

nmps

yxt . From Table A.7, the p-value

associated with t = 3.0 is 2P( t > 3.0 ) = 2(.004) = .008. At significance level .05, Ho is

rejected and we conclude that there is a difference between ∗1µ and ∗

2µ , which is

equivalent to saying that there is a difference between 1µ and 2µ .

b. No. The mean of a lognormal distribution is ( ) 2/2∗∗ += σµµ e , where

∗µ and ∗σ are

the parameters of the lognormal distribution (i.e., the mean and standard deviation of

ln(x)). So when ∗∗ = 21 σσ , then ∗∗ = 21 µµ would imply that 21 µµ = . However,

when ∗∗ ≠ 21 σσ , then even if ∗∗ = 21 µµ , the two means 1µ and 2µ (given by the

formula above) would not be equal. 77. This is paired data, so the paired t test is employed. The relevant hypotheses are

0:0 =dH µ vs. 0: <daH µ , where dµ denotes the difference between the population

average control strength minus the population average heated strength. The observed differences (control – heated) are: -.06, .01, -.02, 0, and -.05. The sample mean and standard

deviation of the differences are 024.−=d and 0305.=ds . The test statistic is

8.176.1024.

50305.

−≈−=−

=t . From Table A.7, with d.f. = 5 – 1 = 4, the lower tailed p-

value associated with t = -1.8 is P( t < -1.8) = P( t > 1.8 ) = .073. At significance level .05, Ho should not be rejected. Therefore, this data does not show that the heated average strength exceeds the average strength for the control population.

261

78. Let 1µ denote the true average ratio for young men and 2µ denote the true average ratio for

elderly men. Assuming both populations from which these samples were taken are normally distributed, the relevant hypotheses are 0: 210 =− µµH vs. 0: 21 >− µµaH . The

value of the test statistic is ( )( ) ( )

5.7

1228.

1322.

71.647.722

=

+

−=t . The d.f. = 20 and the p-value is

P( t > 7.5) 0≈ . Since the p-value is 05.=< α , we reject Ho. We have sufficient evidence to claim that the true average ratio for young men exceeds that for elderly men.

79.

10-1

4

3

2

1

Normal Score

Poo

r Vis

ibili

ty

10-1

2.5

1.5

0.5

Normal Score

Goo

d V

isib

ility

A normal probability plot indicates the data for good visibility does not follow a normal distribution, thus a t-test is not appropriate for this small a sample size.

80. The relevant hypotheses would be FM µµ = versus FM µµ ≠ for both the distress and delight indices. The reported p-value for the test of mean differences on the distress index was less than 0.001. This indicates a statistically significant difference in the mean scores, with the mean score for women being higher. The reported p-value for the test of mean differences on the delight index was > 0.05. This indicates a lack of statistical significance in the difference of delight index scores for men and women.

Chapter 9: Inferences Based on Two Samples

290

81. We wish to test H0: 21 µµ = versus Ha: 21 µµ ≠ Unpooled:

With Ho: 021 =− µµ vs. Ha: 021 ≠− µµ , we will reject Ho if α<− valuep .

( )( ) ( )

1695.15

1113

2

1252.12

1479.

2

1252.1

1479.

22

22

≈=

+

+=ν , and the test statistic

97.14869.

96.36.948.8

1252.1

1479. 22

−=−

=+

−=t leads to a p-value of 2[ P(t > 1.97)]

( ) 062.031.2 ≈≈

Pooled:

The degrees of freedom 24212142 =−+=−== nmν and the pooled variance

is ( ) ( ) 3970.152.12411

79.2413 22 =

+

, so 181.1=ps . The test statistic is

1.2465.

96.

181.1

96.

121

141

−≈−

=+

−=t . The p-value = 2[ P( t24 > 2.1 )] = 2( .023) = .046.

With the pooled method, there are more degrees of freedom, and the p-value is smaller than with the unpooled method.

82. Because of the nature of the data, we will use a paired t test. We obtain the differences by subtracting intake value from expenditure value. We are testing the hypotheses H0: µd = 0 vs

Ha: µd ? 0. Test statistic 88.3757.1

7197.1

==t with df = n – 1 = 6 leads to a p-value of 2[ P( t >

3.88 ) ˜ .004. Using either significance level .05 or .01, we would reject the null hypothesis and conclude that there is a difference between average intake and expenditure. However, at significance level .001, we would not reject.

83.

a. With n denoting the second sample size, the first is m = 3n. We then wish

( )nn

4003900

58.2220 += , which yields n = 47, m = 141.

b. We wish to find the n which minimizes ( )nn

z400

400900

2 2/ +−α , or equivalently, the

n which minimizes nn

400400

900+

−. Taking the derivative with respect to n and

equating to 0 yields ( ) 0400400900 22 =−− −− nn , whence ( )22 40049 nn −= , or

0000,64032005 2 =−+ nn . This yields n = 160, m = 400 – n = 240.

Chapter 9: Inferences Based on Two Samples

291

84. Let p1 = true survival rate at Cο11 ; p2 = true survival rate at Cο30 ; The hypotheses are

0: 210 =− ppH vs. 0: 21 ≠− ppH a . The test statistic is ( )nmqp

ppz

11

21

ˆˆ

ˆˆ

+

−= . With

802.9173ˆ1 ==p , and 927.

110102ˆ 2 ==p , 871.

201175ˆ ==p , 129.ˆ =q .

( )( )( )91.3

0320.125.

129.871.

927.802.

1101

911

−=−

=+

−=z . The p-value =

( ) ( ) 0003.49.391.3 =−Φ<−Φ , so reject Ho at any reasonable level. The two survival rates appear to differ.

85.

a. We test 0: 210 =− µµH vs. 0: 21 ≠− µµaH . Assuming both populations have

normal distributions, the two-sample t test is appropriate. The approximate degrees of

freedom ( )

( ) ( )4.11

110102083.

70325125.

042721.22

2

=+

=ν , so we use df = 11.

437.411,0005. =t , so we reject Ho if 437.4≥t or 437.4−≤t The test statistic is

3.3042721.

68.≈=t , which is not 437.4≥ , so we cannot reject Ho. At significance

level .001, the data does not indicate a difference in true average insulin-binding capacity due to the dosage level.

b. P-value = 2P( t > 3.3) = 2 (.004) = .008 which is > .001.

86. ( ) ( ) ( ) ( )[ ]

41111

ˆ4321

244

233

222

2112

−+++−+−+−+−

=nnnn

SnSnSnSnσ

( ) ( ) ( ) ( ) ( )[ ] 2

4321

244

233

222

2112

41111

ˆ σσσσσ

σ =−+++

−+−+−+−=

nnnnnnnn

E . The estimate for

the given data is ( ) ( ) ( ) ( )[ ]

409.50

1225.112601.76561.174096.15=

+++=

Chapter 9: Inferences Based on Two Samples

292

87. 00 =∆ , 1021 == σσ , d = 1, nn142.14200

==σ , so

−Φ=

142.14645.1

nβ ,

giving =β .9015, .8264, .0294, and .0000 for n = 25, 100, 2500, and 10,000 respectively. If

the si 'µ referred to true average IQ’s resulting from two different conditions, 121 =− µµ

would have little practical significance, yet very large sample sizes would yield statistical significance in this situation.

88. 0: 210 =− µµH is tested against 0: 21 ≠− µµaH using the two-sample t test,

rejecting Ho at level .05 if either 131.215,025. =≥ tt or if 131.2−≤t . With 20.11=x ,

68.21 =s , 79.9=y , 21.32 =s , and m = n = 8, sp = 2.96, and t = .95, so Ho is not rejected. In the situation described, the effect of carpeting would be mixed up with any effects due to the different types of hospitals, so no separate assessment could be made. The experiment should have been designed so that a separate assessment could be obtained (e.g., a randomized block design).

89. 210 : ppH = will be rejected at level α in favor of 21: ppH a > if either

645.105. =≥ zz . With 10.ˆ2500250

1 ==p , 0668.ˆ2500167

2 ==p , and 0834.ˆ =p ,

2.40079.0332.

==z , so Ho is rejected . It appears that a response is more likely for a white

name than for a black name.

90. The computed value of Z is 34.14634

4634−=

+−

=z . A lower tailed test would be

appropriate, so the p-value ( ) 05.0901.34.1 >=−Φ= , so we would not judge the drug to be effective.

Chapter 9: Inferences Based on Two Samples

293

91. a. Let 1µ and 2µ denote the true average weights for operations 1 and 2, respectively. The

relevant hypotheses are 0: 210 =− µµH vs. 0: 21 ≠− µµaH . The value of the

test statistic is ( )

( ) ( )43.6

318083.7

39.17

30672.3011363.4

39.17

3096.9

3097.10

63.141924.140222

−=−

=+

−=

+

−=t .

The d.f. ( )

( ) ( )5.57

2930672.3

29011363.4

318083.722

2

=+

=ν , so use df = 57. 000.257,025. ≈t ,

so we can reject Ho at level .05. The data indicates that there is a significant difference between the true mean weights of the packages for the two operations.

b. 1400: 10 =µH will be tested against 1400: 1 >µaH using a one-sample t test

with test statistic m

s

xt

1

1400−= . With degrees of freedom = 29, we reject Ho if

699.129,05. => tt . The test statistic value is 1.100.224.2140024.1402

3097.10

==−

=t .

Because 1.1 < 1.699, Ho is not rejected. True average weight does not appear to exceed 1400.

92. ( )nm

YXVar 21 λλ+=− and X=1̂λ , Y=2λ̂ ,

nmYnXm

++

=λ̂ , giving

nm

YXZ

λλ ˆˆ +

−= . With 616.1=x and 557.2=y , z = -5.3 and p-value =

( )( ) 0006.3.52 <−Φ , so we would certainly reject 210 : λλ =H in favor of

21: λλ ≠aH .

93. 62.11̂ == xλ , 56.2ˆ2 == yλ , 77.1

ˆˆ21 =+

nmλλ

, and the confidence interval is

( )( ) ( )59.,29.135.94.77.196.194. −−=±−=±−

Chapter 9: Inferences Based on Two Samples

294